Codeforces 520B. Two Buttons

方法一：

可以看出，n大于m，直接输出n-m即可。

否则直接BFS了

#include<bitset>#include<map>#include<vector>#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<stack>#include<queue>#include<set>#define inf 0x3f3f3f3f#define mem(a,x) memset(a,x,sizeof(a))#define F first#define S secondusing namespace std;typedef long long ll;typedef pair<int,int> pii;const int N=123,MOD=1e9+7;int boys[111];int girls[111];int dp[111][111];int vis[600000];struct node{    int x,step;};int main(){    int n,m;scanf("%d%d",&n,&m);    if(n >=m){        printf("%d\n",n-m);return 0;    }    queue<node>Q;    node s ;s.x = n;    s.step = 0;    Q.push(s);    while(!Q.empty()){        node now = Q.front();Q.pop();        if(now.x == m){            printf("%d\n",now.step);            return 0;        }        int lx = now.x*2;        int rx= now.x-1;        node nt;nt.step = now.step +1;        if(lx >0 && lx < m*2 && !vis[lx]){            nt.x = lx;            vis[nt.x] = 1;            Q.push(nt);        }        if(rx >0 && rx < m*2  && !vis[rx]){            nt.x = rx;            vis[nt.x] = 1;            Q.push(nt);        }    }    return 0;}

方法二，

Suppose that at some point we perform two operations of type 1 and then one operation of type 2; but in this case one operation of type 2 and one operation of type 1 would lead to the same result, and the sequence would contain less operations then before. That reasoning implies that in an optimal answer more than one consecutive operation of type 1 is possible only if no operations of type 2 follow, that is, the only situation where it makes sense is when n is smaller than m and we just need to make it large enough. Under this constraint, there is the only correct sequence of moves: if n is smaller than m, we just add 1 until they become equal; else we divide n by 2 if it is even, or add 1 and then divide by 2 if it is odd. The length of this sequence can be found in .

题意是n有两种操作，一个是减一。一个是翻倍。

于是，我们倒过来想，m到n有两种操作，一种是加一，一种是减小一半。

  考虑这种情况，当两个数，例如4 6 ，6 + 1 + 1 然后在 /2 ,花费3 ， 6/2 + 1花费2.

因此，超过一次的+1操作在这种情况下都是多消耗时间的，因此，尽可能的使用2操作，/2.直到不能使用，也就是m <= n的时候。

int main(){    int n,m;scanf("%d%d",&n,&m);    int step =0;    while(n < m){        if(m&1){            m++;        }        else{            m>>=1;        }        step++;    }    printf("%d\n",step+n-m);    return 0;}